I want to prove that if $\dfrac{z_1-z_4}{z_1-z_2} \times \dfrac{z_2-z_3}{z_4-z_3}$ is real, then the four complex numbers are concyclic.
Now I'm aware that this can be done by drawing them up arbitrarily and then observing that we can make use of the fact that a quadrilateral is concyclic iff opposite angles are supplementary.
My question is, how do I really do this in detail without hand waving anything? My issues arise when considering cases. I understand that if that cross ratio is real, then its argument can either be 0 or $\pi$. However, I'm having some trouble understanding how we can possibly have the 0 case.
I can see the outline of the proof by considering the arguments and making use of that geometric property, but beyond that I'm having trouble constructing a proper proof.
Lemma 1: Mappings of the form $z\mapsto w = \dfrac{az+b}{cz+d}$ preserve cross-ratios. I.e., if $w_k=f(z_k)$ for $k=1,2,3,4$; then
$$ \text{cross-ratio}=\frac{z_1-z_4}{z_1-z_2} \times \frac{z_2-z_3}{z_4-z_3}=\frac{w_1-w_4}{w_1-w_2} \times \frac{w_2-w_3}{w_4-w_3}. $$
Lemma 2: Every circle and every straight line can be mapped onto $\mathbb R\cup\{\infty\}$ by a mapping of the form considered in Lemma 1. (Here we consider every line to contain the point $\infty$, which may be mapped either to $\infty$ or to some (finite) real number.)
Lemma 3: The inverse of a mapping of the form considered in Lemma 1 is another mapping of that form; consequently every circle is also the image of $\mathbb R\cup\{\infty\}$ under such a mapping.
So see if you can prove the three lemmas and then use them to get your result.